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Questions tagged [p-groups]

Use with the (group-theory) tag. Refers to questions concerning finite groups of prime power order or infinite p-groups such as Prüfer groups, pro-p-groups, and Tarski monsters. This tag is not for p-adic number systems.

31
votes
1answer
10k views

A $p$-group of order $p^n$ has a normal subgroup of order $p^k$ for each $0\le k \le n$

This is problem 3 from Hungerford's section about the Sylow theorems. I have already read hints saying to use induction and that $p$-groups always have non-trivial centres, but I'm still confused. ...
97
votes
2answers
8k views

More than 99% of groups of order less than 2000 are of order 1024?

In Algebra: Chapter 0, the author made a remark (footnote on page 82), saying that more than 99% of groups of order less than 2000 are of order 1024. Is this for real? How can one deduce this result? ...
18
votes
4answers
10k views

There exists only two groups of order $p^2$ up to isomorphism.

I just proved that any finite group of order $p^2$ for $p$ a prime is abelian. The author now asks to show that there are only two such groups up to isomorphism. The first group I can think of is $G=\...
14
votes
3answers
891 views

Known bounds for the number of groups of a given order.

The number of nonisomorphic groups of order $n$ is usually called $\nu(n)$. I found a very good survey about the values. $\nu(n)$ is completely known absolutely up to $n=2047$, and for many other ...
8
votes
3answers
1k views

Abelian $p$-group with unique subgroup of index $p$

Let $G$ be a finite abelian $p$-group with a unique subgroup $H$ of index $p$. It is a fact that $G$ is cyclic. This can be deduced from the classification theorem for finite abelian groups by writing ...
8
votes
2answers
4k views

If $|G| = p^n$ then $G$ has a subgroup of order $p^m$ for all $0\le m <n.$

Prove that if $|G| = p^n$ then $G$ has a subgroup of order $p^m$ for all $0\le m <n.$ Since $G$ is of prime-power order I know $|Z(G)| \ne e$ so there is an $a\in Z(G)$ with order $p$ such that $p ...
10
votes
1answer
5k views

If $H$ is a proper subgroup of a $p$-group $G$, then $H$ is proper in $N_G(H)$.

Let $H$ be a proper subgroup of $p$-group $G$. Show that the normalizer of $H$ in $G$, denoted $N_G(H)$, is strictly larger than $H$, and that $H$ is contained in a normal subgroup of index $p$. Here'...
9
votes
2answers
3k views

Classify groups of order 27

Let $|G|=27$. Prove that all subgroups of index $3$ are normal. Classify all groups of order $27$. I can do the first one, but the classification is overwhelming. I don't even know where to start. ...
9
votes
2answers
3k views

Characterizations of the $p$-Prüfer group

I'm an undergrad student fairly keen on algebra. Over the different algebra courses I've taken, I've often encountered the so-called $p$-Prüfer group on exercises but somehow never got around to them. ...
8
votes
2answers
816 views

What are central automorphisms used for?

A central automorphism is an automorphism $\theta$ for which $x^{-1}\theta(x)\in Z(G)$ for each $x\in G$. It's not difficult to prove that the set of central automorphisms forms a subgroup of $\...
106
votes
0answers
3k views

Is there a characterization of groups with the property $\forall N\unlhd G,\:\exists H\leq G\text{ s.t. }H\cong G/N$?

A common mistake for beginning group theory students is the belief that a quotient of a group $G$ is necessarily isomorphic to a subgroup of $G$. Is there a characterization of the groups in which ...
14
votes
3answers
8k views

Does every group whose order is a power of a prime p contain an element of order p?

I need to know if every group whose order is a power of a prime $p$ contains an element of order $p$? Should I proceed by picking an element $g$ of the group and proving that there is an element in $\...
5
votes
1answer
2k views

Any irreducible representation of a $p$-group over a field of characteristic $p$ is trivial.

In general, we know that if $G$ is a finite group and $K$ is a field, then $K[G]$ (the group algebra) is semisimple whenever $\operatorname{char}(K)$ does not divide the order of $G$. However, this ...
7
votes
1answer
165 views

Bound on the number of p-groups for fixed exponent

It's well-known that for each prime number $p$ there are exactly two groups of order $p^2$, five of order $p^3$, and fifteen of order $p^4$ (at least when $p>3$). I know that the classification of ...
7
votes
2answers
2k views

Subgroups of order $p$ and $p^{n-1}$ in a group of order $p^n$.

I have a group $G$ of order $p^n$ for $n \ge 1$ and $p$ a prime. I am looking for two specific subgroups within $G$: one of order $p$ and one of order $p^{n-1}$. I don't think I would use the Sylow ...
7
votes
3answers
373 views

Deduce that if $G$ is a finite $p$-group, the number of subgroups of $G$ that are not normal is divisible by $p$

Given: Let $G$ be a group, and let $\mathcal{S}$ be the set of subgroups of $G$. For $g\in G$ and $H\in S$, let $g\cdot H=gHg^{-1}$ Question: Deduce that if $G$ is a finite $p$-group, for some prime $...
5
votes
3answers
183 views

at least one element fixed by all the group

$G$ is a p-group and $S$ is a set that $G$ acts on. p does not divide $|S|$. Why is there at least one element $a\in S$ such that $|O(a)|=1$, or in other words, $G_a=G$? I tried to ask this question ...
4
votes
2answers
4k views

Part of simple proof of nontrivial center in p-group

I'm trying to understand the proof of a Burnside theorem (as stated in Beachy's Abstract Algebra p. 328): Let $p$ be prime number. The center of any $p$-group is nontrivial. Now, In the proof they ...
9
votes
2answers
817 views

A question on $p$-groups, and order of its commutator subgroup.

$\textbf{QUESTION-}$ Let $P$ be a p-group with $|P:Z(P)|\leq p^n$. Show that $|P'| \leq p^{n(n-1)/2}$. If $P=Z(P)$ it is true. Now let $n > 1$, then If I see $P$ as a nilpotent group and ...
2
votes
1answer
1k views

Subgroups and quotient groups of finite abelian $p$-groups

Motivation The fundamental theorem of finite abelian groups gives us a concise description of the isomorphism types of finite abelian $p$-groups $G$ (in the following, $p$ is a fixed prime). The ...
1
vote
0answers
132 views

Subextension of a field with Galois series of subextensions of prime degree

Let $p$ be a prime number, and $E/F$ be a field extension. Suppose $E/F$ has a finite series of subfields $$ F = E_0 < E_1 < \cdots < E_n = E $$ with $E_i / E_{i-1}$ Galois of degree $p$ ...
0
votes
1answer
174 views

Sylow subgroups of soluble groups

Suppose $G \leqslant S_p$ acts transitively on $\{1,...,p\}$ for prime $p$. Let $P \leqslant G$ be a Sylow p-subgroup. Is it true that $G$ is soluble <=> $P \triangleleft G$?
3
votes
1answer
96 views

Let G be $\mathbb Z_p\times\dots\times \mathbb Z_p$ . Find A(G).

Let G be $\underbrace{\mathbb Z_p\times\dots\times \mathbb Z_p}_{n \text{ times}}$. Find $A(G)$. I know that $A(G)\cong GL_n(\mathbb Z_p)$. I prove it by taking $\varphi$ from $A(G)$ and show that ...
10
votes
1answer
1k views

Power of commutator formula

A few people remember a commutator formula of the form $[a,b]^n = (a^{-1} b^{-1})^n (ab)^n c$ where $c$ is a product of only a few commutators (say $n-1$) of them. Here $a,b$ are in a (free) group ...
3
votes
1answer
2k views

Normal subgroups of p-groups

Let $G$ be a group of order $p^\alpha$, where $p$ is prime. If $H\lhd G$, then can we find a normal subgroup of $G/H$ that has order $p$?
8
votes
2answers
291 views

Very generic question about Commutator and Center

Let $G$ be a finite group and $G'$ the commutator group of $G$. What can I say about $G' \cap Z(G)$? Could you be as specific as possible about p-Groups?
10
votes
1answer
1k views

Non-abelian $p$-group; abelian subgroups of index $p$

I'm trying to prove the following problem: (a) Let $G$ be a non-abelian $p$-group with an abelian subgroup of index $p$. Then the number of abelian subgroups of $G$ of index $p$ is either $1$ or $p+...
8
votes
1answer
279 views

Unipotent action of pro-$p$-group

Say $p$ and $\ell$ are distinct prime numbers. Let $G$ be a pro-$p$-group which acts continuously on a finite-dimensional $\mathbb{Q}_\ell$-vector space $V$. Assume that the action of $G$ on $V$ is ...
4
votes
1answer
121 views

Finite Group with $n$-automorphism map

If $G$ is a finite group and $\phi(x) = x^{p+1}$ is an automorphism of $G$ with $order(\phi) |p$ then $G$ is a $p$-group...? If the order of $\phi$ is $1$ then $\phi(x) = x = x^{p+1} = x^px \...
6
votes
1answer
426 views

An infinite $p$-group may not be nilpotent

It is well-known fact that every finite $p$-group $G$ is nilpotent. I am asking to have a counter example when $G$ is infinite $p$-group. Thanks.
5
votes
1answer
781 views

Nonabelian groups of order $p^3$

From a little reading, I know that for $p$ and odd prime, there are two nonabelian groups of order $p^3$, namely the semidirect product of $\mathbb{Z}/(p)\times\mathbb{Z}/(p)$ and $\mathbb{Z}/(p)$, ...
4
votes
2answers
539 views

Frattini Subgroup of p-Groups

Letting $P$ be a $p$-group and $\Phi(P)$ be the Frattini subgroup of $P$ (the intersection of all maximal subgroups), the challenge is "Prove that $P/N$ is elementary abelian implies $\Phi(P)≤N$" (...
4
votes
3answers
3k views

In a group $G$ where $\exists!$ nontrivial, proper subgroup, show that $G$ is cyclic and $\lvert G\rvert=p^2$, for $p$ prime.

Suppose that $G$ is a group that has exactly one nontrivial proper subgroup. Then we have to show that $G$ is cyclic and order of $G$ is $p^2$ where $p$ is prime. I tried as, if $a$ and $b$ two ...
3
votes
1answer
248 views

why is a polycyclic group that is residually finite p-group nilpotent?

I am trying to solve an exercise in D. Robinson's book A Course in the Theory of Groups, which asks me to show that if $G$ is polycyclic and residually finite p-group for infinitely many prime p, then ...
3
votes
2answers
104 views

If $G$ is a group of order $250,000 = 2^4 5^6$, show that $G$ is not simple.

If $G$ is a group of order $250,000 = 2^4 5^6$, show that $G$ is not simple. By the Sylow theorem, we have that the number of $2$-sylow subgroups of $G$ $n_2$ satisfy that $$ n_2 \equiv1\mod2\mbox{ ...
1
vote
0answers
218 views

Orders of Elements in Minimal Generating sets of Abelian p-Groups

I'm looking for as much information about the orders of elements in minimal generating sets of finite abelian $p$-groups as possible. What I really need is complete knowledge about the possible orders ...
6
votes
1answer
349 views

Why are there so many groups (up to isomorphism) of order $p^n$ for $n>2$, especially when compared to groups of similar sized order?

While bounds on the number of isomorphism classes of groups of order $p^n$ where $p$ is prime have been known for quite a while (such as the work of Higman$^{[1]}$ and Sims$^{[2]}$) which give us the ...
4
votes
1answer
164 views

Nilpotency class of normal product of dihedral 2-groups

Pessimistically paraphrasing Polya: “if Jack cannot answer a question, there is an easier question Jack also cannot answer.” Hence I ask: Given positive integers $a,b$ describe the set $$N_{a,b} = \...
4
votes
1answer
105 views

On finite 2-groups that whose center is not cyclic

Let $G$ be a finite 2-group such that $\left|\dfrac{G}{Z(G)}\right|=4$, $Z(G)$ is not cyclic and $Z(G)$ has at least one element of order 4. Then prove that there exists an automorphism $\alpha$ of $...
3
votes
1answer
97 views

Why is the first $p$-adic congruence subgroup a pro-$p$ group?

I am trying to see that $\Gamma_2$, defined as the kernel of the natural surjective map $\text{GL}_2(\mathbb Z_p)\to \text{GL}(\mathbb F_p)$ is a pro-$p$ group. So I'm trying to show that every ...
2
votes
2answers
975 views

$p$-group and normalizer

Here is the question: a) Show that if $p$ is a prime number and $P$ is a $p$-subgroup of a finite group $G$, then $[G:P]=[N_G(P):P]$(mod p), where $N_G(P)$ denotes the normalizer of $P$ in $G$. ...
1
vote
0answers
37 views

Upper central series, semidirect products and a lifting property (argument verification)

Let $G$ be a profinite group, and define the lifting property: $(*_p)$ For every extension $1 \to P \to E \to W \to 1$ where $E$ is finite and $P$ is a $p$-group and for every surjective morphism $f\...
1
vote
1answer
158 views

Divisible abelian $q$-group of finite rank

What does "finite rank" mean in the context of divisible abelian $q$-group? A divisible abelian $q$-group of finite rank is always a Prüfer $q$-group or it can be also a finite product of Prüfer $q$-...
0
votes
2answers
300 views

Proper subgroup of a finite $p$-group

I am a beginner to $p$-groups and have the following question at hand: Let $H$ be a proper subgroup of a finite $p$-group $G$. If $|H|=p^s$ , then there exists a subgroup of order $p^{s+1}$ ...
8
votes
0answers
110 views

Reference request: groups of order $p^4$.

I am looking for a textbook or a paper which include the classification of groups of order $p^4$ ($p$ is prime) using generators and relations. In particular I like to understand which group $G$ "...
5
votes
5answers
2k views

Any group of order $2p$ has a subgroup of order $p~$($p$ being a prime)

I want to show without using Sylow theorem that Any group of order $2p$ has a subgroup of order $p~$($p$ being a prime) My attempt: Since $|G|=2p,$ even $\exists~a\neq e\in G$ such that $a^{-1}=a.$...
4
votes
2answers
716 views

Show that a p-group has a faithful irreducible representation over $\mathbb{C}$ if it has a cyclic center

A p-group is a group of order $p^d$ where p is a prime. If the center has order $p^m$ (since its order must divide the order of the group) then we have a one dimensional faithful irreducible ...
4
votes
1answer
119 views

Random Group of order $4096$ with a center of size $2$

How can I create a random group of order $4096$ with a center of size $2$ ? The algorithm should be able to create every possible group with the given properties in principle. I think the list of ...
3
votes
1answer
272 views

Subgroups of abelian $p$-groups

Let $A$ be an Abelian group of prime power order. It can be expressed as a (unique) direct product of cyclic groups of prime power order: $A = \mathbb{Z}_{p^{n_1}} \times \cdots \times \mathbb{Z}_{...
3
votes
1answer
981 views

Order of automorphism group of a $p$-group is divisible by $p$.

Suppose $G$ is a finite $p$-group (where $p$ is prime), so that $|G|=p^n$ for some positive integer $n\ge 2$. How can we prove that $|\text{Aut}(G)|$ is divisible by $p$? Here $\text{Aut}(G)$ is ...